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Optics & Snell's law

by Anieves
Tags: optics, snell
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Anieves
#1
Jan20-06, 07:24 PM
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I am having A LOT of trouble figuring this one out. . .Some please help.. ok here it goes- So there is a beam of light (with and angle of incidence of 29.9 degrees) and so using snell's law I calculated that the angle of refraction is 21.95 degrees (using n1 as air and n2 as water). The ray of light goes into the water and hits a target. How far is the target from the beam if the target is 3m underwater and the gun shooting the beam is 1.48m above the surface of the water? What equation is available if I need to find distance?
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Doc Al
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Jan20-06, 08:07 PM
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Quote Quote by Anieves
How far is the target from the beam...
I assume this means "how far is the target from the line of the original beam". In other words, the apparent position of the object is along the line of the original in air beam, but the actual position is some distance away from that line due to refraction. Make sense?

To find this distance, draw a careful diagram and use some trig. Indicate the refracted beam, the extended original beam, and the location of the object.
Anieves
#3
Jan20-06, 09:04 PM
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Ok, so there is a beam coming down hitting the surface of the water and 29.9 degrees to the horizontal. (that's the theta i) The beam comes down into the water where it moves away from the normal because n2>n1. The angle of refraction that I found was 21.95(that's theta r). N1 is air=1 and N2 is water=1.333. The object is 3m below the water level and the man holding the beam is 1.48m above the surface level. How far is the object from the man holding the beam?

Doc Al
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Jan20-06, 09:29 PM
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Optics & Snell's law

Quote Quote by Anieves
Ok, so there is a beam coming down hitting the surface of the water and 29.9 degrees to the horizontal. (that's the theta i) The beam comes down into the water where it moves away from the normal because n2>n1.
The refracted beam moves towards the normal.
The angle of refraction that I found was 21.95(that's theta r). N1 is air=1 and N2 is water=1.333. The object is 3m below the water level and the man holding the beam is 1.48m above the surface level. How far is the object from the man holding the beam?
It looks like you've stated the problem a bit more clearly. (A different problem than what I assumed from your initial description in the first post.)

Find the coordinates of the man and the object and use them to find the distance between those two points. For example, the coordinates of the man (assume the light is going left to right, and use the point where the beam hits the water as the origin) would be y = +1.48m. The x coordinate can be found by examining the right triangle whose hypotenuse goes from the man to the origin; use trig to find the bottom leg of the triangle, which will be the x coordinate of the man. One angle in that triangle will be the angle of incidence.

Then do the same thing under the water with the object by find a right triangle with the hypotenuse going from object to origin. This way you'll find the x and y coordinates of the object.

(PS: (1) Post these kinds of questions in this forum, not general physics; (2) Don't start a second thread on the same topic.)


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